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lord [1]
2 years ago
7

C%3A%5C%3A%5C%3A%5C%3A%5C%3A%E2%98%9EQuestion%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%7D%7D%7D" id="TexFormula1" title="{\huge\fcolorbox{blue}{black}{\pink{\:\:\:\:\:\:\:\:\:\:\:☞Question\:\:\:\:\:\:\:\:}}}" alt="{\huge\fcolorbox{blue}{black}{\pink{\:\:\:\:\:\:\:\:\:\:\:☞Question\:\:\:\:\:\:\:\:}}}" align="absmiddle" class="latex-formula">
What is Wave Optics?

Thanks (:​
Physics
1 answer:
kumpel [21]2 years ago
3 0

the branch of optics that studies interference, diffraction, polarization, and other phenomena for which the ray approximation of geometric optics is not valid.

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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
3 years ago
You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas
sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

Weight is given by W = 196 N

We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 \frac{a}{g}=\frac{25}{196}

a=1.2512m/sec^2

Initial velocity is given as 0 so u = 0 m/sec

From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 1.2512\times 3^2=5.6307m

7 0
3 years ago
A merry-go-round with a a radius of R = 1.9 m and moment of inertia I = 209 kg-m2 is spinning with an initial angular speed of ω
RideAnS [48]

Answer:

340.67 kgm²/s

Explanation:

R = Radius of merry-go-round = 1.9 m

I = Moment of inertia = 209 kgm²

\omega_i = Initial angular velocity = 1.63 rad/s

m = Mass of person = 73 kg

v = Velocity = 4.8 m/s

Initial angular momentum is given by

L=I\omega_i\\\Rightarrow L=209\times 1.63\\\Rightarrow L=340.67\ kgm^2/s

The initial angular momentum of the merry-go-round is 340.67 kgm²/s

8 0
3 years ago
The fastest man alive can run the 100 meter dash in 9.58 seconds, calculate average speed in meters per second
densk [106]

Answer:

100 ÷ 9.58 = 10.44 (approximate answer)

3 0
3 years ago
From the center of the Earth to the moon, what should the orbital radius of such satellite be in order to stay over the same poi
yulyashka [42]

In order to have a period that matches the Earth's rotation, a satellite must be in a circular orbit, and 42,164 km from the center of the Earth.

But that's not quite enough to make sure that it always stays over the same point on the Earth's surface (and appears motionless in the sky). For that to happen, the satellite's orbit has to be directly over the Equator.

The Moon has nothing to do with any of this.

3 0
3 years ago
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